Question: $\dfrac{dy}{dx}=\dfrac1x$ and $y(e)=-2$. $y\!\left(e^3\right)=$
Explanation: The differential equation is separable. What does it look like after we separate the variables? $dy=\dfrac1x\,dx$ Let's integrate both sides of the equation. $\int dy=\int\dfrac1x\,dx$ What do we get? $y = \ln |x| + C$ What value of $C$ makes the solution curve pass through the point $(e,-2)$ ? Let's substitute $x=e$ and $y=-2$ into the equation and solve for $C$. $\begin{aligned} -2 &= \ln |e| + C\\ \\ -2&=1+C\\ \\ C&=-3 \end{aligned}$ Now use this value of $C$ to find $y$ when $x=e^3$. $\begin{aligned} y &= \ln\,\left|e^3\right| - 3\\ \\ &=3-3\\ \\ &=0 \end{aligned}$